CBSE Class 12 · 2005 · All India · ElectrostaticsShort Answer

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Define electric field intensity. Write its S.I. unit. Write the magnitude and direction of electric field intensity due to an electric dipole of length

2 a

at the mid-point of the line joining the two charges.

Solution

Electric field intensity is the force per unit positive charge,

E = F / q_{0}

; its S.I. unit is

N C^{- 1}

(or

V m^{- 1}

).
The mid-point M of the dipole is at a distance

a

from each charge.
Field at M due to

+ q

(directed from

+ q

towards

- q

E_{+} = \frac{1}{4 π ε _{0}} \frac{q}{a ^{2}}

Field at M due to

- q

(same direction, equal magnitude):

E_{-} = \frac{1}{4 π ε _{0}} \frac{q}{a ^{2}}

Being collinear and equal, they add:

E = E_{+} + E_{-} = \frac{2 k q}{a ^{2}}

directed from the positive charge towards the negative charge (anti-parallel to

p

CBSE Class 12 Board · 2005 · All India · Subjective

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